What is the $pH$ of the resulting solution when equal volumes of $0.1M$ $NaOH$ and $0.01M$ $HCl$ are mixed ?
A.$7.0$
B.$1.04$
C.$12.65$
D.$2.0$

Hint: To calculate $pH$ first of all we have to find concentration of ${H^ + }$ ion also $pH = - \log \left[ {{H^ + }} \right]$ and to calculate $pOH$ we have to calculate concentration of $O{H^ - }$ ion in the resulting solution . We should remember the formula $pH + pOH = 14$.

Complete step by step answer:
 Chemical reaction for the given solution is : .
Let us assume volume of both $NaOH$ and $HCl$ is equal to $1L$ then we have $0.1$ moles of $NaOH$ and $0.01$ moles of $HCl$ by using the formula $n = M \times V$where $n$ is number of moles , $M$ is molarity and $V$is volume of the solution .
Since we have $0.1$ mole of $NaOH$ and only $0.01$ mole of $HCl$ , so only $0.01$ mole of $O{H^ - }$ ion will take part in neutralisation reaction .
Now number of moles of $O{H^ - }$ ion $ = 0.1 - 0.01 = 0.09$
Total volume of solution $ = 2L$
Concentration of $O{H^ - }$ $ = \dfrac{{0.09}}{2}$ $ = 0.045M$
$pOH = - \log \left[ {O{H^ - }} \right] = - \log \left[ {0.045} \right]$
$pOH = 1.35$
Now we know that $pH + pOH = 14$ . So $pH$ of the resulting solution will be equal to $14 - pOH$ .
$pH = 14 - pOH = 14 - 1.35$
$pH = 12.65$

So option (C) is the correct answer .


Note:
 High $pH$ means that the solution is basic in nature as it is clear from the above question because it has an excess amount of $NaOH$ and its $pH$ value is greater than seven . Similarly the high value of $pOH$ means solution is acidic in nature . Neutral solutions have both $pH$ and $pOH$ of seven . Another point to be noted down is that $pH$ decreases with increase in temperature because as the temperature rises molecular vibrations increases which results in the ability of water to ionise and form more hydrogen ions . As a result $pH$ will drop . But we should not be confused that water becomes more acidic at a higher temperature because a solution is considered to be acidic if there is excess hydrogen ions . In the case of water there is always the same concentration of ${H^ + }$ and $O{H^ - }$.